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Shortcut Trick

By using the Alligation Method on average salaries:

Ratio of Officers to Non-officers = (Overall Avg − Non-officers Avg) : (Officers Avg − Overall Avg)

Ratio = (16000 − 10000) : (28000 − 16000)

Ratio = 6000 : 12000 = 1 : 2

Since 1 ratio part = 15 officers, 2 ratio parts = 15 × 2 = 30 non-officers.

∴ The correct answer is 30.

Alternate Method

Given:

Average salary of officers = ₹28,000

Average salary of non-officers = ₹10,000

Overall average salary = ₹16,000

Number of officers = 15

Officers Non-Officers

28,000 10,000

Overall Avg 16,000

6,000 12,000

Ratio = 1 : 2

Calculations:

Let the number of non-officers be x.

Total Salary = Overall Average × Total Employees

⇒ (15 × 28000) + (x × 10000) = 16000 × (15 + x)

⇒ 420000 + 10000x = 240000 + 16000x

⇒ 16000x − 10000x = 420000 − 240000

⇒ 6000x = 180000

⇒ x = 180000 ÷ 6000

⇒ x = 30

∴ The correct answer is 30.

Additional Information

Weighted Average

When two or more groups are combined, the overall average is calculated as (n₁A₁ + n₂A₂) ÷ (n₁ + n₂), where n is the number of elements and A is the respective average of each group.

Alligation Rule

A rapid visual method to find the ratio in which two quantities are mixed to produce a mixture of known mean. The ratio equals (Dearer − Mean) : (Mean − Cheaper).

Average Deviation Method

The sum of net deviations of all the observations from their exact arithmetic mean is always mathematically equal to zero.

Shortcut Trick

The sum of the three given results is 210 + 198 + 192 = 600.

Each number is added twice as a full unit and once as a half unit across the three equations, giving a combined sum of 2.5 times the total sum of the three numbers.

Sum of the three numbers = 600 ÷ 2.5 = 240.

Required Average = 240 ÷ 3 = 80.

∴ The correct answer is 80.

Alternate Method

Given:

Let the three numbers be a, b, and c.

Calculations:

a + b + c/2=210b + c + a/2=198c + a + b/2=1922.5 × (a + b + c)=600Average = (a + b + c) ÷ 3 = 240 ÷ 3 = 80

According to the given conditions:

⇒ a + b + c/2 = 210

⇒ b + c + a/2 = 198

⇒ c + a + b/2 = 192

Adding all three equations:

⇒ (a + b + c/2) + (b + c + a/2) + (c + a + b/2) = 210 + 198 + 192

⇒ 2(a + b + c) + 0.5(a + b + c) = 600

⇒ 2.5 × (a + b + c) = 600

⇒ a + b + c = 600 ÷ 2.5

⇒ a + b + c = 240

The average of the three numbers is their sum divided by 3:

⇒ Average = (a + b + c) ÷ 3

⇒ Average = 240 ÷ 3 = 80

∴ The correct answer is 80.

Additional Information

Average (Arithmetic Mean)

The average of a given set of numbers is equal to the sum of all the observations divided by the total number of observations. Average = Sum of observations ÷ Total number of observations.

Properties of Average

If each observation in a dataset is increased, decreased, multiplied, or divided by a constant value 'k', the new average will also be correspondingly increased, decreased, multiplied, or divided by 'k'.

Weighted Average

If values x₁, x₂, ..., x_n have associated weights w₁, w₂, ..., w_n, the weighted average is calculated as (w₁x₁ + w₂x₂ + ... + w_nx_n) ÷ (w₁ + w₂ + ... + w_n).

Shortcut Trick

Target questions to pass = 60% of 120 = 72

Current correct questions = (70% of 20) + (50% of 50) + (30% of 50)

⇒ 14 + 25 + 15 = 54

Questions needed to pass = 72 − 54 = 18

∴ The correct answer is 18 questions.

Alternate Method

Given:

Total questions = 120

General Knowledge questions = 20, correctly answered = 70%

Reasoning questions = 50, correctly answered = 50%

English questions = 50, correctly answered = 30%

Minimum passing score = 60% of total questions

Total: 120 questionsCorrect: 5418NeededPassing Score: 72Target = 60% of 120 = 72

Calculations:

⇒ Passing score = 60% of 120 = (60 ÷ 100) × 120 = 72

⇒ Correct GK questions = 70% of 20 = (70 ÷ 100) × 20 = 14

⇒ Correct Reasoning questions = 50% of 50 = (50 ÷ 100) × 50 = 25

⇒ Correct English questions = 30% of 50 = (30 ÷ 100) × 50 = 15

⇒ Total correctly answered questions = 14 + 25 + 15 = 54

⇒ Additional questions needed = Passing score − Total correctly answered

⇒ Additional questions needed = 72 − 54 = 18

∴ The correct answer is 18 questions.

Additional Information

Passing Percentage

The minimum benchmark score set by examination bodies to qualify. Formula: Passing Marks = (Passing % ÷ 100) × Total Marks.

Aggregate Marks

The sum total of marks obtained across all sections or subjects. Formula: Aggregate % = (Total Marks Obtained ÷ Total Maximum Marks) × 100.

Weighted Average

Used when combining groups with different base values or sections with different maximum marks. Formula: Weighted Average = (N₁x₁ + N₂x₂ + ... ) ÷ (N₁ + N₂ + ... ).

Shortcut Trick

Rate for the 1^st year = 9%

Rate for the next 8 months = 9% × (8/12) = 6%

Effective CI rate for 1 year 8 months = 9 + 6 + (9 × 6)/100 = 15.54%

Total Compound Interest = 15.54% of ₹7,000 = ₹1,087.80

∴ The correct answer is ₹1,087.80.

Alternate Method

Given:

Principal (P) = ₹7,000

Rate of Interest (R) = 9% p.a.

Time (T) = 1 year 8 months = 1 + (8/12) years = 1 + (2/3) years

Formula Used:

When time is in a fraction like n + a/b years:

Amount (A) = P × (1 + R/100)ⁿ × [1 + (R × a/b)/100]

Compound Interest (CI) = A − P

Start ₹7,000

End of 1 Year ₹7,630

End of 1 Yr 8 Mo ₹8,087.80

• 9% (₹630)

• 6% (₹457.80) Total CI = ₹630 + ₹457.80 = ₹1,087.80

Calculations:

⇒ Amount (A) = 7000 × (1 + 9/100)¹ × [1 + (9 × 2/3)/100]

⇒ A = 7000 × (109/100) × (1 + 6/100)

⇒ A = 7000 × 1.09 × 1.06

⇒ A = 7000 × 1.1554 = ₹8,087.80

⇒ CI = Amount − Principal

⇒ CI = 8087.80 − 7000 = ₹1,087.80

∴ The correct answer is ₹1,087.80.

Additional Information

Compound Interest Formula

When interest is compounded annually, the total amount is calculated as A = P × (1 + R/100)ⁿ, where P is the principal, R is the rate of interest, and n is the time in years.

Effective Rate of Interest

For two different continuous time periods with rates a% and b%, the combined effective rate is given by the successive percentage change formula: Effective Rate = a + b + (a × b)/100.

Compounding Half-Yearly

When the interest is compounded half-yearly, the rate of interest becomes half (R/2) and the number of time periods becomes double (2n).

Shortcut Trick

In compound interest, the amount from the previous year becomes the principal for the next year.

Interest for the 3^rd year = Amount of 3^rd year − Amount of 2^nd year

⇒ Interest = 13200 − 12000 = ₹1200

Rate of Interest (R) = Interest ÷ Principal of that year

⇒ R = 1200 ÷ 12000 = 0.1

∴ The correct answer is 0.1.

Alternate Method

Year 0 Year 2 Year 3

P = ? ₹12,000 ₹13,200

CI = ₹1,200

Rate = R

Given:

Amount at the end of the 2^nd year (A₂) = ₹12,000

Amount at the end of the 3^rd year (A₃) = ₹13,200

Formula Used:

Amount for n years, A = P(1 + R)ⁿ

Calculations:

⇒ For 2 years: P(1 + R)² = 12000   --- (Equation 1)

⇒ For 3 years: P(1 + R)³ = 13200   --- (Equation 2)

⇒ Dividing Equation 2 by Equation 1:

⇒ [P(1 + R)³] ÷ [P(1 + R)²] = 13200 ÷ 12000

⇒ 1 + R = 1.1

⇒ R = 1.1 − 1

⇒ R = 0.1

∴ The correct answer is 0.1.

Additional Information

Compound Interest Formula

The amount after n years is given by A = P(1 + R)ⁿ, where P is the principal and R is the decimal rate of interest.

Simple vs Compound Interest

For the first year, both Simple Interest (SI) and Compound Interest (CI) are equal if compounded annually on a given principal.

Successive Amounts in CI

The ratio of amounts for any two consecutive years is always constant and equal to the multiplying factor (1 + R).

Shortcut Trick Let the quantities bought be 3, 4, and 3 respectively, making a total of 10 notebooks. Cost of 3 notebooks of Type 1 = 3 × (20 ÷ 5) = ₹12 Cost of 4 notebooks of Type 2 = 4 × (9 ÷ 2) = ₹18 and 3 notebooks of Type 3 = 3 × (45 ÷ 10) = ₹13.5 Total CP of 10 notebooks = 12 + 18 + 13.5 = ₹43.5 Total SP of 10 notebooks = 10 × (19 ÷ 4) = ₹47.5 Profit % = [(47.5 − 43.5) ÷ 43.5] × 100 = (4 ÷ 43.5) × 100 ≈ 9.2% Profit. ∴ The correct answer is Profit of 9.2%.

Alternate Method Given: CP of 5 notebooks of Type 1 = ₹20 CP of 2 notebooks of Type 2 = ₹9 CP of 10 notebooks of Type 3 = ₹45 Mixing ratio of Type 1, Type 2, and Type 3 = 3 : 4 : 3 SP of 4 notebooks = ₹19 Formula Used: Profit = Selling Price (SP) − Cost Price (CP) Profit % = (Profit ÷ CP) × 100

Type 1: 3 CP = ₹12

Type 2: 4 CP = ₹18

Type 3: 3 CP = ₹13.5

Total CP = ₹43.5

Total 10 Notebooks Sold @ ₹19 for 4 SP = 10 × (19 ÷ 4) = ₹47.5

Total SP = ₹47.5

Profit = ₹4

Profit % = (4 ÷ 43.5) × 100 ≈ 9.2%

Calculations: ⇒ Let the number of notebooks of Type 1, Type 2, and Type 3 be 3x, 4x, and 3x respectively. ⇒ Total number of notebooks = 3x + 4x + 3x = 10x ⇒ CP of 1 notebook of Type 1 = 20 ÷ 5 = ₹4 ⇒ Total CP of 3x notebooks of Type 1 = 3x × 4 = ₹12x ⇒ CP of 1 notebook of Type 2 = 9 ÷ 2 = ₹4.5 ⇒ Total CP of 4x notebooks of Type 2 = 4x × 4.5 = ₹18x ⇒ CP of 1 notebook of Type 3 = 45 ÷ 10 = ₹4.5 ⇒ Total CP of 3x notebooks of Type 3 = 3x × 4.5 = ₹13.5x ⇒ Total CP of all 10x notebooks = 12x + 18x + 13.5x = ₹43.5x ⇒ SP of 1 notebook = 19 ÷ 4 = ₹4.75 ⇒ Total SP of all 10x notebooks = 10x × 4.75 = ₹47.5x ⇒ Profit = Total SP − Total CP = 47.5x − 43.5x = ₹4x ⇒ Profit % = (Profit ÷ Total CP) × 100 ⇒ Profit % = (4x ÷ 43.5x) × 100 = 400 ÷ 43.5 = 800 ÷ 87 ≈ 9.195% ⇒ Profit % ≈ 9.2% ∴ The correct answer is Profit of 9.2%.

Additional Information Basic Profit & Loss Rule Profit occurs when Selling Price (SP) > Cost Price (CP). The fundamental formula to calculate the percentage is: Profit % = (Profit ÷ CP) × 100. Mixture Replacement / Combined Cost Price When different quantities or qualities of items are mixed, calculate the per-unit Cost Price of each type first. Then multiply them by their respective assumed ratio amounts to find the combined Total CP, ensuring accurate overall profit/loss estimation.

Shortcut Trick

Let the Selling Prices be 300, 400, and 500 units.

Total SP = 300 + 400 + 500 = 1200 units.

Total CP = (300 ÷ 1.25) + (400 ÷ 1.1) + (500 ÷ 0.8) = 240 + 363.63 + 625 = 1228.63 units.

Net Loss = Total CP − Total SP = 1228.63 − 1200 = 28.63 units.

Loss % = (28.63 ÷ 1228.63) × 100 ≈ 2.33%.

∴ The correct answer is loss of 2.33%.

Alternate Method

Given:

Ratio of Selling Prices (SP) = 3 : 4 : 5

Profit on C1 = 25%, Profit on C2 = 10%, Loss on C3 = 20%

Calculations:

⇒ The fractional values are: 25% = 1/4, 10% = 1/10, 20% = 1/5.

⇒ For C1, CP : SP = 4 : 5

⇒ For C2, CP : SP = 10 : 11

⇒ For C3, CP : SP = 5 : 4

⇒ We need SPs to be in ratio 3 : 4 : 5. First, make SP parts equal by multiplying with LCM(5, 11, 4) = 220.

⇒ Now multiply each SP by its ratio share: SP₁ = 3 × 220 = 660, SP₂ = 4 × 220 = 880, SP₃ = 5 × 220 = 1100.

⇒ Calculate corresponding CPs: CP₁ = 660 × (4 ÷ 5) = 528

⇒ CP₂ = 880 × (10 ÷ 11) = 800

⇒ CP₃ = 1100 × (5 ÷ 4) = 1375

⇒ Total CP = 528 + 800 + 1375 = 2703 units

⇒ Total SP = 660 + 880 + 1100 = 2640 units

Total Cost Price = 2703 units

Total Selling Price = 2640 units

Loss = 63

Transaction Summary

⇒ Loss = Total CP − Total SP = 2703 − 2640 = 63 units

⇒ Loss % = (Loss ÷ Total CP) × 100 = (63 ÷ 2703) × 100 = (2100 ÷ 901)% ≈ 2.33%

∴ The correct answer is loss of 2.33%.

Additional Information

Successive Profit or Loss

When multiple profits/losses are given successively, the effective percentage change is calculated as: Net % = a + b + (a × b) ÷ 100. Use positive sign for profit and negative for loss.

Cost Price from Selling Price

When Profit % is given: CP = (SP × 100) ÷ (100 + P%). When Loss % is given: CP = (SP × 100) ÷ (100 − L%).

Dishonest Dealer Concept

If a dealer claims to sell goods at CP but uses false weight, Profit % = [Error ÷ (True Weight − Error)] × 100.

Shortcut Trick

Let CP = 100 units ⇒ MP = 150 units (50% markup).

Price after 20% discount = 150 − 30 = 120 units.

Overall profit is 8% ⇒ Final SP = 100 + 8 = 108 units.

Second discount (y) reduces price from 120 to 108 ⇒ y = (12 ÷ 120) × 100 = 10%.

108 units = ₹4320 ⇒ 1 unit = 40 ⇒ CP (100 units) = ₹4000.

∴ The correct answer is CP = ₹4000, y = 10%.

Alternate Method

CP = 100u

MP = 150u

SP1 = 120u

SP2 = 108u

+50%

−20%

−y%

Profit = +8%

108u = ₹4320 → 1u = ₹40 → CP (100u) = ₹4000

Given:

Markup = 50%, First discount = 20%, Overall profit = 8%, Final SP = ₹4320.

Calculations:

Let the Cost Price (CP) be x.

⇒ Marked Price (MP) = x + (50% of x) = 1.5x

⇒ Price after 1^st discount = 1.5x × (1 − 20÷100) = 1.5x × 0.8 = 1.2x

⇒ Price after 2^nd discount = 1.2x × (1 − y÷100) = Final SP

⇒ Overall Profit = 8%, so Final SP = x + (8% of x) = 1.08x

Equating both expressions for Final SP:

⇒ 1.2x × (1 − y÷100) = 1.08x

⇒ 1 − y÷100 = 1.08 ÷ 1.2 = 0.9

⇒ y÷100 = 1 − 0.9 = 0.1

⇒ y = 10

Now, calculating CP using the Final SP:

⇒ 1.08x = 4320

⇒ x = 4320 ÷ 1.08 = 4000

⇒ CP = ₹4000

∴ The correct answer is CP = ₹4000, y = 10%.

Additional Information

Successive Discounts

When two discounts d₁% and d₂% are given one after another, the equivalent single discount is given by (d₁ + d₂ − d₁×d₂÷100)%.

Relation between CP and MP

When an item is sold at a profit of P% after offering a discount of D%, the ratio of Cost Price to Marked Price is given directly by: CP ÷ MP = (100 − D) ÷ (100 + P).

Markup Percentage

Markup percentage represents how much higher the marked price is compared to the cost price. It is calculated as: Markup% = (MP − CP) ÷ CP × 100.

Shortcut Trick

Let total work = LCM of 12 and 18 = 36 units.

Efficiencies: A = 36 ÷ 12 = 3 units/day, B = 36 ÷ 18 = 2 units/day.

Work done by A and B together in 4 days = (3 + 2) × 4 = 20 units.

Remaining work = 36 − 20 = 16 units.

Time taken by A to finish the remaining work = 16 ÷ 3 = 5¹/₃ days.

∴ The correct answer is 5¹/₃ days.

Alternate Method

Given:

Time taken by A to complete the task = 12 days

Time taken by B to complete the task = 18 days

Time A and B worked together = 4 days

Formula Used:

Total Work = LCM of individual days

Efficiency = Total Work ÷ Days

Time = Work ÷ Efficiency

Calculations:

⇒ Combined efficiency of A and B = 3 + 2 = 5 units/day.

⇒ Work completed by both in 4 days = 5 × 4 = 20 units.

⇒ Remaining work = 36 − 20 = 16 units.

⇒ Time taken by A for remaining work = Remaining work ÷ Efficiency of A.

⇒ Time taken = 16 ÷ 3 = 5¹/₃ days.

∴ The correct answer is 5¹/₃ days.

Additional Information

Pipes & Cisterns

Functions similarly to Time & Work. Inlets (filling pipes) have positive efficiency, while outlets (emptying pipes) have negative efficiency.

Work & Wages

Wages are distributed in direct proportion to the actual amount of work done by each individual.

Alternate Day Work

Work is done in turns. Calculate the total work done in a complete cycle (e.g., 2 days) and scale it up to reach close to the total work, then find the remainder for the next turn.

Shortcut Trick

By applying the Rule of Alligation directly:

Quantity Ratio = (Mean Price − Price of 2^nd quality) / (Price of 1^st quality − Mean Price)

⇒ 4 / 5 = (130 − x) / (150 − 130)

⇒ 4 / 5 = (130 − x) / 20

⇒ 130 − x = 16 ⇒ x = 114

∴ The correct answer is ₹114.

Alternate Method

Given:

Cost of the 1^st quality tea = ₹150 per kg

Cost of the mixture (at no profit, no loss) = ₹130 per kg

Ratio of 1^st to 2^nd quality mixed = 4 : 5

Formula Used:

Total Cost = (Quantity₁ × Price₁) + (Quantity₂ × Price₂)

₹150 1^st Quality

₹x 2^nd Quality

130 Mixture Price

4 parts 130 − x

5 parts 150 − 130 = 20

Calculations:

Let the quantities of the 1^st and 2^nd quality tea be 4k and 5k respectively.

Let the cost of the 2^nd quality tea be ₹x per kg.

⇒ Total quantity of the mixture = 4k + 5k = 9k

⇒ Total cost of the mixture = 9k × 130 = 1170k

⇒ Total cost of the 1^st quality = 4k × 150 = 600k

⇒ Total cost of the 2^nd quality = 5k × x = 5kx

Since the total cost of individual parts must equal the total mixture cost:

⇒ 600k + 5kx = 1170k

Dividing the entire equation by k:

⇒ 600 + 5x = 1170

⇒ 5x = 1170 − 600

⇒ 5x = 570

⇒ x = 570 ÷ 5 = 114

∴ The correct answer is ��114.

Additional Information

Rule of Alligation

It allows us to directly find the ratio in which two components at given prices must be mixed to obtain a specific mean price. Ratio = (Cheaper Quantity) : (Dearer Quantity) = (D − M) : (M − C).

Mean Price

The effective cost price of a unit quantity of the combined mixture is known as the mean price. It always lies strictly between the cost prices of the cheaper and dearer components being mixed.

Weighted Average Logic

Alligation is an alternative derivation of the weighted average formula: A_w = (w₁x₁ + w₂x₂) ÷ (w₁ + w₂), where w represents weights (quantities) and x represents values (prices).